Problem: We know that $\frac{1}{1+x^2}=1-x^2+{{x}^{4}}-{{x}^{6}}+...$ for $x\in\left(-1,1\right)$. Using this fact, find the function that corresponds to the following series. $ x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}-\frac{x^7}{7}+...$ Choose 1 answer: Choose 1 answer: (Choice A) A $\arctan(-x)$ (Choice B) B $\arctan(x)$ (Choice C) C $\arctan(x^2)$ (Choice D) D $\ln(x)$ (Choice E) E $\ln(-x)$ (Choice F) F $-\ln(x)$
First, note that the derivative of $ x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}-\frac{x^7}{7}+...$ is $ 1-x^2+x^4-x^6-...=\frac{1}{1+x^2}$ Then $\int{\frac{1}{1+x^2}}~dx=x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}-\frac{x^7}{7}+...$ That is, $\arctan(x) + C=x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}-\frac{x^7}{7}+...$ Now let $x=0\,$ ; since $\arctan(0)=0\,$, we know that $~C=0$. Thus, $ x-\frac{{{x}^{3}}}{3}+\frac{{{x}^{5}}}{5}-\frac{x^7}{7}+...=\arctan(x)$